Question

Am nevoie de ajutor pentru problema aceasta de la UTCN!!!

Answer 1

$f(x) = \sqrt[5]{x^3-\tan ^3 x} = (x^3-\tan^3 x)^{\frac{1}{5}}\\ \\ \\ f'(0) = \lim\limits_{x\to 0}\dfrac{f(x) - f(0)}{x-0} =\lim\limits_{x\to 0}\dfrac{(x^3-\tan^3 x)^{\frac{1}{5}}}{x} = \\ \\ =\lim\limits_{x\to 0} \Big(\dfrac{x^3-\tan^3 x}{x^5}\Big)^{\frac{1}{5}} = \Big( \lim\limits_{x\to 0} \dfrac{x^3-\tan^3 x}{x^5}\Big)^{\frac{1}{5}}$

$\lim\limits_{x\to 0} \dfrac{x^3-\tan^3 x}{x^5} = \lim\limits_{x\to 0} \dfrac{3x^2-3\tan^2 x(\tan^2 x+1)}{5x^4} = \\ \\ = \dfrac{3}{5}\lim\limits_{x\to 0} \dfrac{x^2-\tan^4 x-\tan^2 x}{x^4} = \\ \\ = \dfrac{3}{5}\Big(\lim\limits_{x\to 0} \dfrac{x^2-\tan^2 x}{x^4} -\lim\limits_{x\to 0} \dfrac{\tan^4 x}{x^4}\Big) = \\ \\ = \dfrac{3}{5}\Big(\lim\limits_{x\to 0} \dfrac{x^2-\tan^2 x}{x^4} -1\Big) =$

$= \dfrac{3}{5}\Big(\lim\limits_{x\to 0}\dfrac{2x-2\tan x(\tan^2 x+1)}{4x^3}-1\Big) \\ \\ =\dfrac{3}{10}\Big(\lim\limits_{x\to 0}\dfrac{x-\tan^3 x-\tan x}{x^3}-2\Big) = \\ \\ = \dfrac{3}{10}\Big(\lim\limits_{x\to 0}\dfrac{x-\tan x}{x^3}-3\Big) = \\ \\ = \dfrac{3}{10}\Big(\lim\limits_{x\to 0}\dfrac{1-\tan^2x-1}{3x^2}-3\Big) = \\ \\ =\dfrac{3}{10}\Big(-\dfrac{1}{3}-3\Big)= \\ \\ = \dfrac{3}{10}\cdot \Big(-\dfrac{10}{3}\Big) = \\ \\ = -1$

$\Rightarrow f'(0) = (-1)^{\frac{1}{5}} \Rightarrow \boxed{f'(0) = -1}$